Hot-air balloon (kinematics question)

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A hot-air balloon rises at a constant velocity of 4.0 m/s before a can of pop is dropped from a height of 4.0 m. The key question is whether the can starts with an initial velocity of 0 m/s or 4.0 m/s due to the balloon's motion. The relevant equation for displacement under constant acceleration is d(t) = d0 + v0 t + (1/2) a t^2, where gravity acts as a negative acceleration of -9.8 m/s². The discussion emphasizes the importance of correctly identifying initial conditions and values to solve for the time it takes for the can to reach the ground. Properly applying the equation with the correct initial velocity and displacement will yield the solution.
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Homework Statement



A hot-air balloon is rising upward with a constant velocity of 4.0 m/s. As the balloon reaches a height of 4.0 m above the ground, the balloonist accidentally drops a can of pop over the edge of the basket. How long does it take for the pop can to reach the ground?

Homework Equations



d = v2t - 1/2at^2
other uniform acceleration equations

The Attempt at a Solution



Tried a few different things, but nothing seems to work out. I'm not sure if the pop can starts at v1 = 0 m/s or at v1 = 4.0 m/s because of the rising balloon. It falls from 4.0 m and gravity is -9.8 m/s^2.

That's 4 values, but which of the 4 do I use? I didn't get the right answer from just plugging them in, so I'm not sure what else I can do.
 
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Sean1218 said:

Homework Statement



A hot-air balloon is rising upward with a constant velocity of 4.0 m/s. As the balloon reaches a height of 4.0 m above the ground, the balloonist accidentally drops a can of pop over the edge of the basket. How long does it take for the pop can to reach the ground?

Homework Equations



d = v2t - 1/2at^2

Just to eliminate potential problems, in the velocity term, do you mean v2 t, (v2)t, or something else? The equation is either incorrect as written or at least confusing.
This should be an equation for displacement d(t) as a function of:

1. time, t
2. displacement at t=0, d0
3. velocity at t=0, v0
4. acceleration, a.

Try to find the appropriate equation. It's close to the one that you wrote.

other uniform acceleration equations

The Attempt at a Solution



Tried a few different things, but nothing seems to work out. I'm not sure if the pop can starts at v1 = 0 m/s or at v1 = 4.0 m/s because of the rising balloon. It falls from 4.0 m and gravity is -9.8 m/s^2.

That's 4 values, but which of the 4 do I use? I didn't get the right answer from just plugging them in, so I'm not sure what else I can do.

What was the velocity of the can before it was dropped?
 
Sorry, meant v2 as in velocity #2.

d = v#2(t) - 1/2at^2 where v#2 = 0, d = 4, a = -9.8, solve for t

if I do it with v#1 instead (v#1 = 4.0), I still don't seem to get the right answer. I must just be plugging in the wrong values, but I've tried everything I can think of.

and the velocity of it before it was dropped was 4.0 m/s, so I guess that'd be v#1, yea.
 
I told you to find a more general equation, since you're having trouble identifying how to plug the information given into the equation that you're using. Find the most general equation for displacement at constant acceleration (in your notes or text) and identify the initial and final values that you need to put in.
 
i thought that was what I did :confused:
 
It's not, the most general equation is

d(t) = d0 + v0 t + (1/2) a t2,

where

d(t) is the displacement after the time t
d0 is the displacement at t=0
v0 is the velocity at t=0
a is the constant acceleration.

It's conventional to measure displacement as increasing in the vertical direction. The signs of the initial velocity and acceleration must be identified consistent with that convention.

Try identifying all of the quantities in the equation above in terms of the information given.
 
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